Nee help in PL/SQL or SQL

Question

We need to know that how many SID are getting created and deleted,existing for each time stamp : -
You can assume that for the first timestamp entry, all SID are new.
for example : -
TIMESTAMP SID
11/12/2008 1:25:02 PM 10
11/12/2008 1:25:02 PM 20 Total SID Count = 3
11/12/2008 1:25:02 PM 30
11/12/2008 1:30:02 PM 10 Total SID Count = 2
11/12/2008 1:30:02 PM 40
11/12/2008 1:35:00 PM 40
11/12/2008 1:35:00 PM 50 Total SID Count = 4
11/12/2008 1:35:00 PM 60
11/12/2008 1:35:00 PM 70
from above we can conclude that
I want this output
TIMESTAMP New SID Existing SID Deleted SID
11/12/2008 1:25:02 PM 3 (10,20 and 30 are new) 0 0
11/12/2008 1:30:02 PM 1 (40 is new) 1 (10 is existing) 2 (20 and 30 are deleted)
11/12/2008 1:35:00 PM 3 (50,60 and 70 are new) 1 (40 is existing) 1 (10 is deleted)

Answer 1

Hi,
It's easy to accidentally post the same thread twice.
If you do, edit one of the copies. Change the subject line to "Duplicate - Ignore" and, if the thread has already been answered, post a link to the other copy.
This query does what you requested:
WITH     g     AS
(     -- Begin sub-query g to compute grp_num (1, 2, 3, ...)
     SELECT     tmstmp
     ,     sid
     ,     DENSE_RANK () OVER (ORDER BY tmstmp)     AS grp_num
     FROM     table_x
)     -- End sub-query g to compute grp_num (1, 2, 3, ...)
,     sj     AS
(     -- Begin sub-query sj to outer-join each row to its predecessor
     SELECT     t.tmstmp
     ,     t.sid
     ,     CASE
               WHEN  p.tmstmp IS NULL     THEN 'N'
                              ELSE 'E'
          END     AS status_cd
     FROM          g     t
     LEFT OUTER JOIN     g     p     ON     t.sid          = p.sid
                         AND     t.grp_num     = p.grp_num + 1
)     -- End sub-query sj to outer-join each row to its predecessor
,     c     AS
(     -- Begin sub-query c to compute COUNTs (except deleted_sid)
     SELECT     tmstmp
     ,     COUNT (*)                         AS total_sid
     ,     COUNT (CASE WHEN status_cd = 'N' THEN 1 END)     AS new_sid
     ,     COUNT (CASE WHEN status_cd = 'E' THEN 1 END)     AS existing_sid
     FROM     sj
     GROUP BY     tmstmp
)     -- End sub-query c to compute COUNTs (except deleted_sid)
SELECT     tmstmp
,     new_sid
,     existing_sid
,     new_sid
     + LAG (total_sid, 1, 0) OVER (ORDER BY tmstmp)
     - total_sid     AS deleted_sid
FROM     c
ORDER BY     tmstmp
;You can tell if a sid is existing or new by doing a self-join, where you outer-join each row to the same sid and the previous tmstmp (timestamp is the name of a data type, so it's not a good name for a column). If a row has a match in the previous tmstmp, then it's existing. If a row doesn't have a match, then it's new. Sub-query sj, above, does this.
To determine which is the previous timestmp, you can first assign each distinct tmstmp a sequential number (1, 2, 3, ...) using the DENSE_RANK analytic function. Sub-query g, above, does this.
Getting the count of new and existing sids at each tmstmp is just a matter of doing an aggregate COUNT. Sub-query c, above, does this.
The number of deletions at any tmstmp is equal to difference between
(a) the number of new sids plus the total sid's from the previous tmstmp, and
(b) the total sids from the current tmstmp
The main query, above, does this, using the analytic LAG function.